Postgraduate Diploma Dissertation

The Abdus Salam International Centre for Theoretical Physics Trieste, Italy Postgraduate Diploma Dissertation Title Syzygies Author Yairon Cid Ruiz Supervisor Prof. Lothar Göttsche - August, 2016 -

ii To mom and dad, of course.

Algebra is the offer made by the devil to the mathematician. The devil says: I will give you this powerful machine, it will answer any question you like. All you need to do is give me your soul: give up geometry and you will have this marvellous machine. Michael Atiyah iii

Acknowledgements I am very grateful to my supervisor Lothar Göttsche for his invaluable support, guidance and encouragement throughout the preparation of this project and during all this year in ICTP. Very special thanks go to Tarig Abdelgadir for his advices, the long and helpful conversations, and all the interesting things I was able to learn in his reading courses. I feel deep gratitude in my heart to all the professors of the Mathematics Section of ICTP for showing me a whole new world and for helping me to better appreciate the beauty of mathematics. Finally, I thank the people responsible of my well-being. I thank my beloved Elizabeth and my family for their love, patience and support during this stormy year. iv

Contents Introduction 1 1 Basic concepts 3 1.1 Modules and Resolutions............................ 3 1.2 Two important types of rings......................... 5 1.3 Homology and derived functors........................ 8 1.4 Regular sequences and the Koszul complex.................. 13 1.5 Dimension and depth.............................. 23 2 Some fundamental theorems on syzygies 31 2.1 Towards Hilbert s syzygy theorem and more................. 31 2.2 On Serre s conjecture.............................. 36 2.2.1 A theorem of Serre........................... 37 2.2.2 The Calculus of Unimodular Rows.................. 44 2.3 The Auslander-Buchsbaum formula...................... 49 3 Applying syzygies 53 3.1 The µ-basis of a rational surface........................ 53 3.2 The freeness of the syzygies.......................... 58 3.3 The outline of our proof............................ 61 3.4 Bounds for the regularity and the Betti numbers............... 64 3.5 The case with projective dimension one.................... 72 3.6 The case with projective dimension two.................... 74 3.7 Appendix.................................... 79 Bibliography 82 v

Introduction The word syzygy comes to mathematics from a rather interesting route. It comes from the Greek syzygos and meaning yoked together. It was widely used in astronomy in the 18th century to describe three planets in a line (and thus yoked by some mysterious force). Cayley and Sylvester were the first to use it in mathematics. The current usage inside mathematics is as follows. If M is a module and F a free module with a surjective map ϕ : F M, then the kernel K = Ker(ϕ) is called the 0th syzygy of M. The three modules K, F and M lie in a line 0 K F ϕ M 0, and one can view this as K being yoked to M. In general we can continue this process of finding the syzygies and construct a free resolution for a module. For example, let R = K[x, y] and M = R/m where m = (x, y). A generator for M is given by 1 and we find 0 m R π M 0. The module of syzygies is m, which is generated by x and y. These elements are not independent, since they satisfy the non-trivial relation ( y)x + xy = 0. Resolving again for m we get y ( ) x x y 0 R R 2 m 0. Finally gluing together these two exact sequences we find a free resolution for M y ( ) x x y 0 R R 2 R π M 0. 1

2 It turns out that the finiteness of the process in the previous example is not casual and every finitely generated module M (not necessarily graded) over a polynomial ring R = K[x 1,..., x n ], where K is a field, has a free resolution of length at most n. As in many places of mathematics Hilbert was the first person looking at this type of problems; and in his famous Syzygy Theorem, published in 1890, he states that every finitely generated graded module M = M i over the polynomial ring C[x 1,..., x n ] has a free resolution of length at most n. In this report we shall study several results and theorems related with syzygies, just like previous one. The outline is as follows In Chapter 1, we summarize the basic concepts and tools coming from Commutative Algebra and Homological Algebra that we will need throughout this work. In Chapter 2, we shall present three important and beautiful theorems The Hilbert s Syzygy Theorem which could be seen as the starting point of this now rich theory. The Quillen-Suslin Theorem that answered positively the Serre s conjecture: every finitely generated projective module over K[x 1,..., x n ] is free. The Auslander-Buchsbaum formula which is a surprising relation between the depth and the projective dimension of modules over a local ring. In Chapter 3, we accomplish two objectives at the same time We give answer to a question posed by David Cox in the paper [1]. Along the path of answering this question, we present a short introduction to the theory in the case of graded rings. One of the main goals of this report is giving a fair introduction to the theory of syzygies and make it an accessible read for any graduate student without previous knowledge on the subject (like myself).

Chapter 1 Basic concepts In this chapter we will introduce some basic facts with the hope of making a self-contained work, but we will only prove a statement, if proving it is as easy as quoting it. The main reference for the topics related with Commutative Algebra will be the classic book by Atiyah and MacDonald [10] and for the use of some tools coming from Homological Algebra we will follow the book by Rotman [19]. Throughout all this exposition the word ring shall mean a commutative ring with an identity element, hence we will differ mainly in this aspect from the text [19], e.g. for us will not exist left R-modules or right R-modules, for us only matters the term R-modules. 1.1 Modules and Resolutions Definition 1.1. Let R be a ring (commutative with identity element, as always) then an R-module is an abelian group M on which R acts linearly by a mapping R M M that satisfies the axioms a(x + y) = ax + ay (a + b)x = ax + bx (ab)x = a(bx) 1x = x for all a, b R and x, y M. 3

1.1. Modules and Resolutions 4 One of the most natural ways of constructing an R-module is taking several copies of R, more formally by defining the free module which is an R-module of the form i I M i, where each M i = R. It turns out that we can always give a good description of an arbitrary module using these free modules. Construction 1.2. Let M be an arbitrary module and choose a set of generators {x i } i I of this module, then we define the free module F 0 = i I Rx i ; hence we can obtain an d exact sequence 0 K 0 F 0 0 M 0. We may repeat this operation for K0 to obtain the exact sequence 0 K 1 F 1 K 0 0, and iterating successively we get 0 K n F n K n 1 0 where each F n is a free module. Gluing all these short exact sequences we obtain the long exact sequence... d n+1 d F n d n... 2 d 1 d F1 0 F0 M 0 which is called a free resolution of M. The intermediate modules K n = Ker(d n ) = Im(d n+1 ) are called the syzygies of M. After this construction several questions come to our minds because certainly this construction is not unique neither the syzygies of M. To solve this problem of uniqueness we will use some techniques from Homological Algebra, the path that we have chosen to achieve this goal is different from the one taken in the book by Eisenbud [3]. We have decided to work in the general settlement of a commutative ring with a unit element without assuming extra properties, although later on we will use this marvelous text [3] for stronger results that can only be achieved making additional assumptions. The main ingredient for this generalization will be that of a projective module. Definition 1.3. A projective module P is a module that given any surjective R-linear map β : B C and any R-linear map α : P C, there exists an R-linear map φ : P B such that α = β φ, i.e makes the following diagram φ P α B C 0 β commute.

1.2. Two important types of rings 5 This type of module will give us the fundamental tool for describing modules in general. Definition 1.4. A projective resolution of a module M is an exact sequence... d n+1 d P n d n... 2 d 1 d P1 0 P0 M 0 in which each P n is a projective module. The study of projective modules generalizes that of free modules because every free module is a projective module, and we will see that by doing this we have identified the key property that will abstract and simplify our work. Theorem 1.5. Every free module is a projective module. Proof. (See [19], page 61 ) Then we define the term syzygy for any projective resolution of a module and we will see how this theory is independent of the chosen projective resolution. Definition 1.6. Let... d n+1 d P n d n... 2 d 1 d P1 0 P0 M 0 be a projective resolution of M. For n 0 we denote K n = Ker(d n ) as the nth syzygy of M. 1.2 Two important types of rings In this short section we shall introduce the basic definitions of modules over a local ring and over a graded polynomial ring. One of the biggest advantages of working over these two types of rings is the possibility of applying Nakayama s lemma in a pleasant way. Definition 1.7. A ring R with exactly one maximal ideal m is called a local ring and will be denoted by (R, m) (or (R, m, K) if we want to stress the residue field K = R/m). Definition 1.8. For the polynomial ring R = K[x 1,..., x n ] we will use the standard grading with deg(x c 1 1 x c 2 2... x cn n ) = c 1 + c 2 +... + c n. The K-vector space generated by the monomials of degree i is denoted by R i. The graded polynomial ring R has a direct sum decomposition R = i N R i as k-vector spaces with R i R j R i+j for all i, j N.

1.2. Two important types of rings 6 We say that M is a graded R-module, if it has a direct sum decomposition M = i Z M i as K-vector spaces and R i M j M i+j for all i, j Z. Of particular interest for us is the free graded module R( p) defined by R( p) i = R i p, i.e., for p 0 the module R( p) is shifted p degrees with R( p) p = R 0 = K. Lemma 1.9. (Nakayama s lemma for local rings) Let (R, m, K) be a local ring and M be a finitely generated R-module. (i) If mm = M, then M = 0. (ii) If N is a submodule of M and M = mm + N, then M = N. Proof. (i) Suppose that {x 1,..., x r } is a minimal system of generators for M. Using the hypothesis mm = M we have x 1 = α 1 x 1 +... + α r x r for some α 1,..., α r m. Since 1 α 1 is invertible we get the contradiction x 1 = (1 α 1 ) 1 (α 2 x 2 +... + α r x r ). (ii) Apply (i) to the module M/N. Lemma 1.10. (Nakayama s lemma for graded rings) Let R = K[x 1,..., x n ] be a graded polynomial ring, M be a graded finitely generated R-module, and I R be a homogeneous proper ideal. (i) If IM = M, then M = 0. (ii) If N is a submodule of M and M = IM + N, then M = N. Proof. (i) Since M is finitely generated there exists a smallest integer j Z with M j 0. Because I is a proper ideal all homogeneous elements in I has positive degree and this implies the contradiction M j IM. (ii) Apply (i) to the module M/N. With the use of Nakayama s lemma we can define the concept of minimal number of generators for finitely generated modules over a local ring or over a graded polynomial ring, i.e., all minimal systems of generators have the same number of elements. Lemma 1.11. Let (R, m, K) be a local ring (R = K[x 1,..., x n ] be the graded polynomial ring with m = (x 1,..., x n ) the irrelevant ideal). Let M be a finitely generated R-module,

1.2. Two important types of rings 7 then the minimal number of generators is equal to µ(m) = dim K (M R K). Proof. We shall prove that {v 1,..., v r } is a minimal system of generators for M if and only if the set of residue classes {v 1,..., v r } is a basis for the K-vector space M R K. If v 1,..., v r M/mM are a K-basis of M R K then we have that M = Rv 1 +...+Rv r +mm. Then Nakayama s lemma ((ii) in Lemma 1.9 or (ii) in Lemma 1.10) implies that M = Rv 1 +... + Rv r. If v 1,..., v r is a minimal system of generators for M, then the residue classes v 1,..., v r M/mM( = M R K) clearly are a generating set for M R K, but we claim they also conform K-basis. By contradiction suppose there is a linear dependence between v 1,..., v r, therefore there exists a strictly smaller subset {v i1,..., v is } {v 1,..., v r }, such that {v i1,..., v is } is a K-basis for M R K. By the previous implication we get the contradiction that {v i1,..., v is } is a generating set for M. Theorem 1.12. Let (R, m, K) be a local ring (R = K[x 1,..., x n ] be the graded polynomial ring with m = (x 1,..., x n ) the irrelevant ideal). Suppose F ψ G ϕ H 0 is an exact sequence of R-modules, with G a free module. Then the elementary vectors of G are mapped into a minimal system of generators of H if and only if Im(ψ) mg. Proof. From the initial exact sequence we can get the exact sequence F/mF ψ G/mG ϕ H/mH 0. The map ϕ is an isomorphism if and only if ϕ maps the elementary vectors of G/mG into a basis of the K-vector space H/mH ( = H R K), by Nakayama s lemma this happens if and only if ϕ maps the elementary vectors of G into a minimal system of generators of H. The map ϕ is an isomorphism if and only if Im(ψ) = 0, that is Im(ψ) mg.

1.3. Homology and derived functors 8 1.3 Homology and derived functors Definition 1.13. A complex (or chain complex) C is a sequence of modules and maps with d n d n+1 = 0 for all n Z. d n+1 d C :... A n+1 A n n An 1..., n Z The condition d n d n+1 = 0 is equivalent to Im(d n+1 ) Ker(d n ) and therefore every exact sequence could be seen as a chain complex. Here our main construction will be a functor H n : COMP M R from the category of chain complexes [14] into the category R-modules. Definition 1.14. If C is a complex, its nth homology module is H n (C) = Ker(d n )/Im(d n+1 ). Given a projective resolution of the module M we construct the deleted complex by suppressing M and obtaining the complex... d n+1 d P n d n... 2 d 1 P1 P0 0 and by doing this we do not loose any information since M = Coker(d 1 ) = P 0 /Im(d 1 ) = P 0 /Ker(d 0 ). Then we will apply certain functors to this complex and by computing the homology of the resulting complex we will obtain the derived functors T or and Ext. That will be our two fundamental tools. d n+1 Definition 1.15. Given a module M with its respective deleted complex C :... d n d... 2 d 1 P1 P0 0 coming from a projective resolution. For any module N we P n apply the functor R N D :... d n+1 R 1 P n R N dn R1... d 2 R 1 P 1 R N d 1 R 1 P 0 R N 0 and we define Tor R n (M, N) = H n (D) = Ker(d n R 1)/Im(d n+1 R 1). d n+1 Definition 1.16. Given a module M with its respective deleted complex C :... d n d... 2 d 1 P1 P0 0 coming from a projective resolution. For any module N we P n

1.3. Homology and derived functors 9 apply the functor Hom R (, N) E : 0 Hom R (P 0, N) d 1 Hom R (P 1, N) d 2... d n Hom R (P n, N) d n+1... and we define Ext n R(M, N) = H n (E) = Ker(d n+1)/im(d n). Now surprisingly enough we have that both functors are independent of the projective resolution chosen which is a result coming from the property of projective modules (see [19], Theorem 6.9 (Comparison Theorem)). Theorem 1.17. Let M and N be R-modules then (1) T orn R (M, N) = T orn R (N, M). (2) T or0 R (M, N) = M R N. (3) T orn R (M, N) does not depend on the projective resolution chosen. Proof. (See [19], Chapter 8 ). Theorem 1.18. Let M and N be R-modules then (1) Ext 0 R(M, N) = Hom R (M, N). (2) Ext n R(M, N) does not depend on the projective resolution chosen. (3) For all n we have Ext n R( M k, N) = Ext n R(M k, N). Proof. (See [19], Chapter 7 ). A fundamental result that we will use several times is the following long exact sequence argument for the functors Tor and Ext. Theorem 1.19. Suppose 0 A B C 0 is a short exact sequence of R-modules. Then we have the long exact sequence... T orn R (A, N) T orn R (B, N) T orn R (C, N)...... T or1 R (A, N) T or1 R (B, N) T or1 R (C, N) T or0 R (A, N) T or0 R (B, N) T or0 R (C, N) 0, the long exact sequence

1.3. Homology and derived functors 10 0 Ext 0 R(C, N) Ext 0 R(B, N) Ext 0 R(A, N) Ext 1 R(C, N) Ext 1 R(B, N) Ext 1 R(A, N)...... Ext n R(C, N) Ext n R(B, N) Ext n R(A, N)..., and also the exact sequence 0 Ext 0 R(N, A) Ext 0 R(N, B) Ext 0 R(N, C) Ext 1 R(N, A) Ext 1 R(N, B) Ext 1 R(N, C)...... Ext n R(N, A) Ext n R(N, B) Ext n R(N, C).... Proof. (See [19], Theorem 6.21, Theorem 6.26 and Theorem 6.27 ). Using the concept of split exact sequence (for a good explanation on this, see [13] page 131 ) there is a complete characterization of projective modules. Theorem 1.20. A module M is projective if and only if every short exact sequence 0 A B M 0 splits. Proof. ( ) First suppose M is projective then over an exact sequence 0 A B M 0 we can construct the following diagram φ M id 0 A B M 0, where the induced map φ makes the sequence split. i d ( ) For the reverse we only need that the sequence 0 K 0 F 0 0 M 0 splits then we get F 0 = M K0. We know that F 0 is projective and now M can be completely identified inside F 0 by a projection and an inclusion map. Consider the diagram

1.3. Homology and derived functors 11 ω F 0 π M i φ M β B C 0, α then we see that M satisfy diagram of Definition 1.3 with φ = ω i, and finally M is a projective module. Corollary 1.21. A module M is projective if and only if M is a direct summand of a free R-module. More specifically, M is projective if and only if F 0 = M K0. Proof. Look at the previous proof and see that all is around the exact sequence 0 K 0 d F 0 0 M 0. Example 1.22. We have Z/15Z = Z/5Z Z/3Z, then both Z/5Z and Z/3Z are projective Z/15Z-modules, but neither of them are free because they have less than 15 elements. And we end this section with the following theorem that is a stone where much of our work rests. We have tried to give our own proof by using explicitly the syzygies of the construction 1.2, which seems reasonable given the title of this work. i Theorem 1.23. Let M be a module if Ext 1 (M, N) = 0 for every module N, then M is a projective module. Proof. We compute the 0th syzygy of M 0 K 0 i F 0 d 0 M 0, (1.1) then we glue it with 1.2 to get the following commutative diagram d n+1 d n d 3 d 2... F n... F 2 F 1 F 0 M 0 d 1 d 1 id id i 0 K 0 F 0 M 0 d 0 d 0

1.3. Homology and derived functors 12 where d 1 is the restriction of d 1 to Im(d 1 ) = K 0 and i denotes the inclusion map. By hypothesis we have 0 = Ext 1 (M, K 0 ) = Ker(d 2)/Im(d 1), that implies Ker(d 2) = Im(d 1). Hence we will try to relate this property on the first row of the previous commutative diagram to a property on the second row. We will obtain that 1.1 is a split exact sequence. First lets translate the module Ker(d 2) in terms of 1.1: Ker(d 2) = {f Hom(F 1, K 0 ) (f d 2 )(x) = 0 x F 2 } = {f Hom(F 1, K 0 ) f Im(d2 )= 0} = {f Hom(F 1, K 0 ) f K1 = 0}, the quotient K 0 = F 1 /K 1 induces an isomorphism Ker(d 2) = Hom(K 0, K 0 ) 1 given by the map Φ : Hom(K 0, K 0 ) Ker(d 2), where Φ(ϕ)(x) = ϕ([x]). This map is well-defined from the R-linearity of ϕ( ) and [ ], is an R-linear map because Φ(α ϕ+β ψ)(x) = (α ϕ+β ψ)([x]) = α ϕ([x])+β ψ([x]) = α Φ(ϕ)(x)+β Φ(ψ)(x). It is bijective because any f Hom(F 1, K 0 ) that vanishes on K 1 is constant on every coset of F 1 /K 1. From the previous commutative diagram we take the following square d 1 F 1 F 0 d 1 i K 0 F 0 id and applying the Hom(, K 0 ) functor we get the commutative diagram d 1 Hom(F 1, K 0 ) Hom(F 0, K 0 ) d 1 id i Hom(K 0, K 0 ) Hom(F 0, K 0 ). 1 A good way of imagine this isomorphism is like when we compute the coordinate ring of an affine variety.

1.4. Regular sequences and the Koszul complex 13 Now the isomorphism Hom(K 0, K 0 ) = Ker(d 2) = Im(d 1 : Hom(F 0, K 0 ) Hom(F 1, K 0 )) gives the surjection Hom(F 0, K 0 ) Hom(K 0, K 0 ) 0, which we identify in the previous diagram to obtain 0 Hom(K 0, K 0 ) Hom(F 0, K 0 ) ϕ id i Hom(K 0, K 0 ) Hom(F 0, K 0 ), where ϕ is an isomorphism, because the fact that d 1 is surjective implies that d 1 is injective. From this follows that necessarily the map i : Hom(F 0, K 0 ) Hom(K 0, K 0 ) is surjective and there exists g Hom(F 0, K 0 ) such that i (g) = id K0, that is g i = id K0. Therefore we say that the exact sequence 1.1 splits and following the proof of the previous Theorem 1.20 we get that M is projective. Remark 1.24. For a short proof of the previous theorem, see [19] page 199, simply apply Theorem 1.19 to the short exact sequence 1.1 and obtain Hom(F 0, K 0 ) i Hom(K 0, K 0 ) Ext 1 (M, K 0 ) = 0. 1.4 Regular sequences and the Koszul complex In this section we will discuss the concept of regular sequence and its tight relation with the Koszul complex. Our exposition will be mixed between the books [9], [11] and [6]. From [9] we will use Section 4 in Chapter 21, from [11] the Section 16 and from [6] the Section 1.6. Definition 1.25. Let R be a ring and M an R-module. An element x R is regular in M if xm 0 for any nonzero m M. A sequence of elements x 1,..., x n in R is called M-regular (or just M-sequence) if the two following conditions hold (i) M/(x 1,..., x n )M 0 (ii) x 1 is regular in M, and for any i > 1, x i is regular in M/(x 1,..., x i 1 )M. The sequence (x 1,..., x n ) is called a weak M-sequence if only the condition (ii) is required to be satisfied.

1.4. Regular sequences and the Koszul complex 14 Theorem 1.26. If x 1,..., x n is an M-sequence then then so is x k 1 1,..., x kn n for any positive integers k 1,..., k n. Proof. See [11], Theorem 16.1. We start with a small discussion on the simplest type of Koszul complex. For any x R we can construct K(x) : 0 R x R 0, and here we see how special are the homologies, with H 1 (K(x)) = Ann(x) and H 0 (K(x)) = R/xR. More generally, for elements x 1,..., x n R we define the Koszul complex K(x) = K(x 1,..., x n ) as follows. For x = (x 1,..., x n ) the modules of the complex K(x) will be K 0 (x) = R; K 1 (x) = free module E with basis {e 1,..., e n };. K p (x) = free module p E with basis ei1... e ip, i 1 <... < i p ;. K n (x) = free module n E of rank 1 with basis e1... e n. The boundary map is defined by d(e i ) = x i and in general d : K p (x) K p 1 (x) by p d(e i1... e ip ) = ( 1) j+1 x ij e i1... ê ij... e ip. j=1 A simple verification shows that d 2 = 0, then using the previous modules and the boundary map, we define K(x) as K(x) : 0 K n (x)... K 1 (x) K 0 (x) 0. In the previous definition we see that the zero homology is H 0 (K(x)) = R/I, where

1.4. Regular sequences and the Koszul complex 15 I = (x 1,..., x n ). Then we define the augmented Koszul complex as 0 K n (x)... K 1 (x) K 0 (x) R/I 0. The next Lemma gives the invariant property that we would like for the Koszul complex. Suppose that x = (x 1,..., x n ) and y = (y 1,..., y n ) are n-tuples of elements in R. If the ideal I = (y 1,..., y n ) is contained in the ideal I = (x 1,..., x n ), then we can make n y i = a ij x j with a ij R. j=1 Let {e 1,..., e n} be a basis for K 1 (y) then we define the R-linear map f : K 1 (y) K 1 (x) given by f(e i) = that we can extend for any p (1 p n) as n a ij e j, j=1 f p : K p (y) K p (x), where f p = f... f (p-times). In particular, we know that f n is just the multiplication by the determinant of the matrix A = (a ij ) t of change of basis. Lemma 1.27. With notation as above, the homomorphisms f p define a morphism of augmented Koszul complexes 0 K n (y)... K p (y)... K 1 (y) R R/I 0 det(a) f p f id can 0 K n (x)... K p (x)... K 1 (x) R R/I 0 and defines an isomorphism if det(a) is a unit in R. Proof. See [9], page 853, Lemma 4.2. Therefore, if two n-tuples x = (x 1,..., x n ) and y = (y 1,..., y n ) generate the same ideal and the determinant of the linear transformation is a unit, then the two Koszul complexes are isomorphic. An important and special case is when (y) is a permutation of (x).

1.4. Regular sequences and the Koszul complex 16 Given two complexes K :... e i+1 K i e i Ki 1 e i 1 e... 1 K0 0 and L :... f i+1 L i f i Li 1 f i 1 f... 1 L0 0, its tensor product is defined as the complex K L :... d i+1 j+k=i with boundary map defined by (K j R L k ) d i j+k=i 1 (K j R L k ) d k 1... d 1 (K 0 R L 0 ) 0, d i+j (u R v) = e i (u) R v + ( 1) i u R f j (v) for any u K i and v L j. The change of sign is necessary to make d i+1 d i = 0. Theorem 1.28. For Koszul complexes there is a natural isomorphism K(x 1,..., x n ) = K(x 1 )... K(x n ). Proof. Follows from the definitions of tensor product of complexes and Koszul complex. Now we can extend the notion of Koszul complex for any R-module. Definition 1.29. Let M be an R-module, then we define the Koszul complex of M by K(x; M) = K(x 1,..., x n ; M) = K(x 1,..., x n ) R M that looks like K(x; M) : 0 K n (x) R M... K 1 (x) R M K 0 (x) R M 0. Generally we always have the equalities H 0 (K(x; M)) = M/IM and H n (K(x); M) = {v M x 1 v =... = x n v = 0}. Given an element x R we study what happens what when we tensor an arbitrary complex C :... d p+1 d p d C p... 2 d 1 C1 C0 0 with the simple Koszul complex K(x). We can

1.4. Regular sequences and the Koszul complex 17 make the exact sequence of complexes 0 C C K(x) (C K(x))/C 0. The Koszul complex K(x) is simply 0 R x R 0, then (C K(x)) p = (C p R R) (C p 1 R R) = C p C p 1. Making this identification we can make explicit the previous exact sequence of complexes as i 0 C p+1 C p+1 C p C p 0 d p+1 e p+1 i 0 C p C p C p 1 C p 1 0 d p e p i 0 C p 1 C p 1 C p 2 C p 2 0 π Cp π Cp 1 π Cp 2 d p d p 1 where the boundary map of the complex C K(x) is defined by (i) e p (u) = d p (u) for any u C p ; (ii) e p (v) = d p 1 (v) + ( 1) p 1 xv for any v C p 1. The corresponding homology long exact sequence is given by... H p (C) H p (C K(x)) H p ((C K(x))/C) H p 1 (C)..., by the previous discussions we have H p ((C K(x))/C) = H p 1 (C) and from a simple diagram chasing we get = ( 1) p 1 x. Therefore we get the exact sequence... H p (C) H p (C K(x)) H p 1 (C) ( 1)p 1 x H p 1 (C)..., that can be reordered as... H p (C) ( 1)p x H p (C) H p (C K(x)) H p 1 (C)....

1.4. Regular sequences and the Koszul complex 18 Lemma 1.30. Let x R and let C be a complex as above. Then x annihilates H p (C K(x)) for all p 0. Proof. For p = 0 we have (C K(x)) 0 = C0, then e 1 (u) = xu for any u C 0 and follows xh 0 (C K(x)) = 0. For p 1, if u+v C p C p 1 is a cycle (i.e. e p (u+v) = 0) with u C p and v C p 1, then we get ( 1) p xv = d p (u) and d p 1 (v) = 0. Thus e p+1 (( 1) p u) = ( 1) 2p xu + ( 1) p d p (u) = x(u + v), and we get at once xh p (C K(x)) = 0. We want to apply the previous considerations of the tensor product with C = K(x 1,..., x n 1 ; M) and x = x n. To abbreviate a little bit, we will use the notation H p K(x; M) instead of H p (K(x; M)), and we are going to call it the Koszul homology. Theorem 1.31. Let R be a ring, M an R-module and (x 1,..., x n ) an arbitrary sequence then (i) There is an exact sequence... H p K(x 1,..., x n 1 ; M) ( 1)p x H p K(x 1,..., x n 1 ; M) H p K(x 1,..., x n ; M) H p 1 K(x 1,..., x n 1 ; M) ( 1)p 1 x H p 1 K(x 1,..., x n 1 ; M) H p 1 K(x 1,..., x n ; M)... ; (ii) Every element of I = (x 1,..., x n ) annihilates H p K(x 1,..., x n ; M) for all p 0; (iii) If I = R, then H p K(x 1,..., x n ; M) = 0 for all p 0. Proof. These are consequences from Lemma 1.30 and Theorem 1.28. For an R-module M and a sequence x = (x 1,..., x n ) we can define the augmented Koszul of M with respect to x as 0 K n (x; M)... K p (x; M)... K 1 (x; M) K 0 (x; M) M/IM 0, and after some identifications it has the form 0 M... M (n p)... M n M M/IM 0. Theorem 1.32. Let M be an R-module.

1.4. Regular sequences and the Koszul complex 19 (i) Let x = (x 1,..., x n ) be a regular sequence for M. Then H p K(x; M) = 0 for p > 0 (of course, H 0 K(x; M) = M/IM), i.e., the augmented Koszul complex is exact. (ii) Conversely, suppose R is a local ring (or a graded polynomial ring), and x = (x 1,..., x n ) inside the maximal ideal (or x = (x 1,..., x n ) inside the irrelevant ideal of the polynomial ring). Suppose M is finitely generated over R, and H 1 K(x; M) = 0. Then (x 1,..., x n ) is an M-regular sequence. Proof. (i) We proceed making induction on n. For n = 1 is clear, because as previously seen H 1 K(x; M) = {m M xm = 0} = 0 when x R is regular on M. So we assume n > 1. The case p > 1 comes directly from the piece of exact sequence H p K(x 1,..., x n 1 ; M) H p K(x 1,..., x n ; M) H p 1 K(x 1,..., x n 1 ; M), which from the inductive hypothesis turns into 0 H p K(x 1,..., x n ; M) 0. For p = 1, we take the tail of the sequence of Theorem 1.31 and the isomorphism H 0 K(x 1,..., x n 1 ; M) = M/(x 1,..., x n 1 )M to get the exact sequence H 1 K(x 1,..., x n 1 ; M) H 1 K(x 1,..., x n ; M) M/(x 1,..., x n 1 )M xn M/(x 1,..., x n 1 )M. The multiplication by x n is an injective map because x n is regular on M/(x 1,..., x n 1 )M. From the inductive hypothesis H 1 K(x 1,..., x n 1 ; M) = 0, then joining this two facts we get H 1 K(x 1,..., x n ; M) = 0. (ii) Inductively we are going to prove that H 1 K(x 1,..., x j ; M) = 0 for j = 1,..., n. Initially we have H 1 K(x 1,..., x n ; M) = 0. From Theorem 1.31 we have the exact sequence H 1 K(x 1,..., x j 1 ; M) x j H 1 K(x 1,..., x j 1 ; M) H 1 K(x 1,..., x j ; M), so by the inductive hypothesis H 1 K(x 1,..., x j ; M) = 0 and the multiplication by x j is a surjective map. Then Nakayama s lemma (Lemma 1.9 or Lemma 1.10) gives that necessarily H 1 K(x 1,..., x j 1 ; M) = 0. So H 1 K(x 1,..., x j ; M) = 0 for j = 1,..., n, and from 0 = H 1 K(x 1,..., x j ; M) M/(x 1,..., x j 1 )M x j M/(x 1,..., x j 1 )M we have that the multiplication by x j is injective, i.e., x j is regular on M/(x 1,..., x j 1 )M. To finalize, M/IM 0 from Nakayama s lemma. The previous theorem gives a complete characterization of M-regular sequences in the

1.4. Regular sequences and the Koszul complex 20 case of R is a local ring or a graded polynomial ring, and can be achieved with only checking the simple condition H 1 K(x; M) = 0. Also we can get important results for regular sequences by using the Koszul complex, for example from Lemma 1.27 we know that any permutation of a regular sequence is also regular in the case of local rings or graded polynomial rings. An important case where we want to apply the previous theorem is when M = R, i.e. x = (x 1,..., x n ) is an R-regular sequence. In this case we know that the augmented Koszul complex is 0 K n (x)... K 1 (x) K 0 (x) R/I 0 and looks like 0 R... R (n p)... R n R R/I 0. Therefore we can regard the augmented Koszul complex as a free resolution of R/I, when the defining sequence of I is regular. Definition 1.33. Let M be an R-module and x 1,..., x n be elements in an ideal I. We say that x 1,..., x n is a maximal M-sequence in I if x 1,..., x n is M-regular and x 1,..., x n, y is not M-regular for any y I. When the ring R is a Noetherian any ideal has a maximal M-regular sequence. Surprisingly enough, the length of a maximal M-regular sequence is well-determined by means of the Koszul complex. First we have to prove that the Koszul complex is an exact functor. Proposition 1.34. Let R be a ring, x = (x 1,..., x n ) a sequence in R, and 0 L M N 0 an exact sequence of R-modules. Then the induced sequence 0 K(x; L) K(x; M) K(x; N) 0 is an exact sequence of complexes. Therefore, one has a long exact sequence... H p K(x; L) H p K(x; M) H p K(x; N) H p 1 K(x; L)... of homology modules. Proof. The modules in the Koszul complex are free, hence flat R-modules.

1.4. Regular sequences and the Koszul complex 21 In general the Koszul complex cannot tell you if a sequence is exact or not (in the local and the graded case we saw it is true). But when the ring R is Noetherian, it can give you something even more important. Theorem 1.35. Let R be a Noetherian ring, I = (y 1,..., y n ) an ideal of R, and M a finitely generated R-module such that M IM. If we set q = sup{i H i K(y 1,..., y n ; M) 0} then any maximal M-sequence in I has length n q. Proof. Let x 1,..., x m be a maximal M-sequence in I. We will proceed by induction on m. For m = 0, we have that I consists of zero divisors of M, so there exists an associated prime ideal p Ass R (M) such that I p (See Proposition 1.2.1, [6]). By definition of associated primes, p = Ann R (u) for some u M, hence Iu = 0. Therefore we get u H n K(y 1,..., y n ; M) = {v M Iv = 0} = 0 and in this case the assertion q = n holds. Now suppose m 1, then we take M 1 = M/x 1 M and the exact sequence 0 M x 1 M M 1 0. Applying Proposition 1.34 we get the long exact sequence... H p K(y; M) x 1 H p K(y; M) H p K(y; M 1 ) H p 1 K(y; M) x 1 H p 1 K(y; M)..., using Theorem 1.31 (ii) we get the short exact sequence 0 H p K(y; M) H p K(y; M 1 ) H p 1 K(y; M) 0 for every p. Thus H q+1 K(y; M 1 ) 0 and H p K(y; M 1 ) = 0 for p > q + 1. But x 2,..., x m is a maximal M 1 -sequence in I and by the inductive hypothesis m 1 = n (q + 1). Therefore m = n q. This previous theorem has the interesting interpretation, that regular sequences have always length smaller or equal than the number of elements in any generating set of an ideal and that all maximal regular sequences have the same length. There is a dual version of the Koszul complex from which we can obtain all the same

1.4. Regular sequences and the Koszul complex 22 results. In fact, in the book of Eisenbud [3] is taken this approach to the Koszul cohomology. Definition 1.36. For a sequence x = (x 1,..., x n ) R and an R-module M, the dual Koszul complex is defined by Hom R (K(x), M) Hom R (K(x), M) : 0 Hom R (K 0 (x), M) d 1... d n Hom R (K n (x), M) 0. The Koszul cohomology is given by H p (Hom R (K(x), M)) and we are going to denote it by H p K(x; M). It turns out that the Koszul complex is self-dual. With the isomorphism Hom R (K(x), M) = K(x) R M, we can reduce the problem to prove that K(x) and K(x) are isomorphic. For the dual Koszul complex the basis of each module K p (x) is given by (e i1... e ip ), i 1 <... < i p, where (e i1... e ip ) is the function that takes 1 on e i1... e ip basis elements. and 0 on all the other There is a natural isomorphism ω n : K n (x) R that takes e 1... e n into 1. We can define ω p : K p (x) K n p (x) by setting (ω p (u))(v) = ω n (u v) for u K p (x), v K n p (x). We can consider the following commutative diagram, that in fact induces an isomorphism between the complexes K(x) and K(x) (See [6], pages 47 and 48 ). d n d p+1 d p d 1 K(x) : 0 K n (x)... K p (x)... K 0 (x) 0 ω n ω p ω 0 K(x) : 0 K 0 (x) d 1... d n p d Kn p n p+1... d n K n (x) 0 Theorem 1.37. Let x = (x 1,..., x n ) be a sequence in R. Then (i) The complexes K(x) and K(x) are isomorphic (the previous diagram is an isomorphism between them; we say that K(x) is self-dual). (ii) More generally, for every R-module M the complexes K(x; M) = K(x) R M and

1.5. Dimension and depth 23 Hom R (K(x), M) are isomorphic. (iii) H p K(x; M) = H n p K(x; M) for p = 0,..., n. Proof. See Proposition 1.6.10, [6]. 1.5 Dimension and depth One very basic notion in mathematics is that of dimension and we shall deal with the concept in this section. Here we will call the classic Krull dimension, define the projective dimension and give a meaning to the length of maximal regular sequences. Let R be a ring. The supremum of the lengths r, taken over all strictly decreasing chains p 0 p 1... p r of prime ideals of R, is called the Krull dimension, or simply the dimension of R, and denoted by dim R. The codimension of a prime ideal p is defined as codim p = dim R p and the dimension as dim p = dim R/p. For an arbitrary ideal I we define the codimension as codim I = inf{dim R p p V (I)}. If M is an R-module then the dimension is given by dim M = dim(r/ann(m)). If M is finitely generated then dim M is the combinatorial dimension of the closed subspace Supp(M) = V (Ann(M)) of Spec(R), i.e., the length of the longest chain of closed irreducible subsets in Supp(M). Now comes the numerical invariant with biggest importance for us. Definition 1.38. Let M be an R-module then we define pd R (M) as the projective dimension, where pd R (M) = n if n is the smallest natural number such that there is a projective resolution d 0 P n d n... 2 d 1 d P1 0 P0 M 0.

1.5. Dimension and depth 24 If no such resolution exists then we define pd R (M) =. Example 1.39. pd R (M) = 0 if and only if M is projective. Example 1.40. pd R (R[x 1, x 2,..., x n ]) = 0, because is free with basis {x 1, x 2,..., x n }. Example 1.41. Let V be a module over the field K (vector space) then pd K (V ) = 0, since we can always find a basis. Here we get an important characterization of the projective dimension of a module using the functor Ext. Proposition 1.42. Let {K n } be the syzygies of M coming from a projective resolution of M, then Ext n+1 (M, N) = Ext 1 (K n 1, N) for every module N and every n 1. d Proof. Suppose that... P n d n... 2 d 1 d P1 0 P0 M 0 is the projective resolution that defines the syzygies {K n }. Then using K n 1 = Ker(d n 1 ) = Im(d n ) we get the following projective resolution of K n 1... d n+3 d n+2 d n+1 d P n+2 P n+1 P n n Kn 1 0. Then we delete K n 1 and apply the functor Hom(, N) to obtain 0 Hom(P n, N) d n+1 Hom(P n+1, N) d n+2 Hom(P n+2, N) d n+3... and from this follows Ext 1 (K n 1, N) = Ker(d n+2)/im(d n+1) = Ext n+1 (M, N). Theorem 1.43. The following three conditions are equivalent for a module M (1) pd(m) n; (2) Ext k (M, N) = 0 for all modules N and all k n + 1; (3) Ext n+1 (M, N) = 0 for all modules N. Proof. (1) (2) If pd(m) n then there is a projective resolution where P k = 0 for all k n + 1. Therefore Hom(P k, N) = 0 for any module N and also Ext k (M, N) = Ker(d k+1 : Hom(P k, N) Hom(P k+1, N)) Im(d k : Hom(P k 1, N) Hom(P k, N)) = 0.

1.5. Dimension and depth 25 (2) (3) It is clear. (3) (1) We proceed with the construction 1.2 of finding a free resolution (which is also projective) until we obtain an exact sequence d n 1 0 K n 1 F n 1... d 2 d F 1 d 1 0 F0 M 0 where K n 1 is the (n 1)th syzygy. If we obtain before a zero syzygy then the result follows trivially. Otherwise, by hypothesis and Proposition 1.42 we have Ext 1 (K n 1, N) = Ext n+1 (M, N) = 0 for all N. Then Theorem 1.23 implies that K n 1 is projective and pd(m) n. Corollary 1.44. Let M be an R-module with pd(m) n, then for any projective resolution of M the corresponding syzygies K j (j n 1) are projective modules. Proof. For j n 1 and any module N we have Ext 1 (K j, N) = Ext j+2 (M, N) = 0 by Proposition 1.42 and the previous Theorem 1.43. Thus K j is projective from Theorem 1.23. Corollary 1.45. Suppose M is a module with pd(m) = n < then the sequence d n 1 0 K n 1 F n 1... d 2 d F 1 d 1 0 F0 M 0 from the step n of 1.2 is a projective resolution of M. This previous corollary says that any module M with n = pd R (M) < has a projective resolution of length n that is almost free and that our initial and rather naive construction 1.2 will always give a projective resolution exactly at the step n = pd(m), i.e., a projective resolution of minimal length. Later we will see that for a finitely generated module over K[x 1,..., x n ] we can get a free resolution of length n = pd(m). For a very nice treatment on Free Resolutions one can see [15] in Chapter 6. Corollary 1.46. pd(m) = inf{k N Ext k+1 (M, N) = 0 for all modules N}. Corollary 1.47. Given a family of modules {A i : i I}, then pd( i I A i ) = sup{pd(a i ) : i I}.

1.5. Dimension and depth 26 Proof. Using the previous characterization of projective dimension and Theorem 1.18 (3), the result follows. Example 1.48. Let R = K[x]/(x 2 ) and M the R-module defined by M = (x), then pd R (M) =. 2 Proof. First let s compute who are F 0 and K 0, since M is generated by one element we get F 0 = R, then K 0 = Ker(R x M) = (x) = M. Inductively for any n we get the exact sequence 0 K n R x K n 1 (= M) 0, then all syzygies are equal to Ker(R x M) = M. i For any y R \ M we have xy / R \ M, then the exact sequence 0 K 0 R x M 0 does not split because i(k 0 ) = K 0 (= M) cannot be a direct summand of R. From Theorem 1.20 we get M is not projective. Therefore all syzygies are not projective and by Corollary 1.45 we get pd R (M) =. After defining the projective dimension of any R-module then we can define the global dimension of the ring R. Definition 1.49. The global dimension of a ring R is defined as gdim(r) = sup{pd R (M) M M R }, where M R is the category of R-modules. Example 1.50. gdim(k) = 0 because we always have pd K (V ) = 0. Example 1.51. gdim(k[x]/(x 2 )) = because as previously stated pd K[x]/(x 2 )((x)) =. From Theorem 1.35 in the previous section, we saw that for a Noetherian ring R and a finite R-module M in the case IM M we can define the numeric invariant of length of maximal sequences. Definition 1.52. Let R be a Noetherian ring, M a finite R-module, and I an ideal such that IM M. The common length of all the maximal M-sequences in I is called the depth of I on M, denoted by depth(i, M). When M = R is simply called the depth of I and denoted by depth(i). If IM = M we adopt the convention depth(i, M) =. 2 Taken from [15], Exercise 11, page 258.

1.5. Dimension and depth 27 We now want to give a characterization of depth in terms of the Ext functor. Proposition 1.53. Let R be a ring, and M, N R-modules. Set I = Ann(N). (i) If I contains an M-regular element, then Hom R (N, M) = 0. (ii) Conversely, if R is Noetherian, and M, N are finite and Hom R (N, M) = 0, implies that I contains an M-regular element. Proof. See Proposition 1.2.3, [6]. Proposition 1.54. Let R be a ring, M, N be R-modules, and x = (x 1,..., x n ) a weak M-sequence in Ann(N). Then Hom R (N, M/xM) = Ext n R(N, M). Proof. We proceed by induction on n, for n = 0 we have trivially Hom R (N, M) = Ext 0 R(N, M). Let n 1, and set x = (x 1,..., x n 1 ). By the induction hypothesis we have Ext n 1 R (N, M) = Hom R (N, M/x M). Since x n is (M/x M)-regular, then Ext n 1 R (N, M) = 0 by the previous Proposition 1.53. The exact sequence 0 M x 1 M M/x 1 M 0 yields the exact sequence 0 Ext n 1 R (N, M/x 1 M) Ext n R(N, M) x 1 Ext n R(N, M). But the multiplication by x 1 annihilates N, therefore we have the isomorphism Ext n R(N, M) = Ext n 1 R (N, M/x 1 M). Again, the inductive hypothesis and the fact that (x 2,..., x n ) is regular in M/x 1 M, gives the expected result Ext n R(N, M) = Hom R (N, M/xM). Theorem 1.55. (Rees). Let R be a Noetherian ring, M a finite R-module, and I an ideal such that IM M. Then all maximal M-sequences in I have the same length n given by depth(i, M) = min{i Ext i R(R/I, M) 0}. Proof. Suppose x = (x 1,..., x n ) is a maximal M-sequence in I. From Proposition 1.54 we know that Ext i 1 R (R/I, M) = Hom R (R/I, M/(x 1,..., x i 1 )M) for i = 1,..., n, and

1.5. Dimension and depth 28 by Proposition 1.53 Ext i 1 R (R/I, M) = Hom R (R/I, M/(x 1,..., x i 1 )M) = 0. Since IM M and x is maximal in I we have that I consists of zero-divisors of M/xM, then I is contained in some associated prime p = Ann(m) for some m M/xM (See Proposition 1.2.1 [6]). The assignment 1 m induces a monomorphism φ M/xM, and thus a non-zero homomorphism φ : R/I M/xM. Therefore we have : R/p Ext n R(R/I) = Hom R (R/I, M/xM) 0. A very special case is when (R, m, K) is a Noetherian local ring, and M finite R-module. We call the depth of M as depth(m) = min{i Ext i R(K, M) 0}, where K = R/m is the residue field. At this point we should stress that the condition IM M is not superfluous and the concept of depth is not well defined when IM = M. Example 1.56. Let K be a field and R = K[[x]][y]. Then we have that {x, y} and {xy 1} are both maximal R-sequences. 3 Proof. Since R/(x, y) = K, adding any other q R \ (x, y) will make (x, y, q) = R, thus (x, y) is a maximal R-sequence. Taking the quotient K[[x]][y]/(xy 1) (the Rabinowski s trick ), is like adjoining the inverse of x, i.e., K[[x]][y]/(xy 1) = K[[x]][x 1 ]. In general we know that k=0 a k x k K[[x]] is invertible if and only if a 0 0. Now let 0 k=n a k x k K[[x]][x 1 ], where n Z is the smallest integer with a n 0, then k=n a k x k = (x n )( k=n a k x k n ) is clearly invertible. Therefore K[[x]][y]/(xy 1) is a field and as before {xy 1} is a maximal R-sequence. In the following proposition we collect some important results. (From now on, V (I) denotes the set of prime ideals containing I.) 3 Exercise 1.2.20, [6].

1.5. Dimension and depth 29 Proposition 1.57. Let R be a Noetherian ring, I, J ideals of R, and M a finite R-module. Then (i) depth(i, M) = inf{depth(m p ) p V (I)}; (ii) depth(i, M) = depth(rad(i), M); (iii) depth(i J, M) = min(depth(i, M), depth(j, M)); (iv) if x = (x 1,..., x n ) is an M-sequence in I, then depth(i/(x), M/xM) = depth(i, M/xM) = depth(i, M) n; (v) if N is a finite R-module with Supp(N) = V (I), then depth(i, M) = inf{i Ext i R(N, M) 0}. Proof. See Proposition 1.2.10, [6]. In the case of Noetherian local rings there is an important relation between the depth and the dimension. Proposition 1.58. Let (R, m) be a Noetherian local ring and M 0 a finite R-module. Then depth(m) dim(m). Proof. See Proposition 1.2.12, [6]. Also one could see Theorem 6.5, [11]. This local property leads to the following very desirable inequality. Proposition 1.59. Let R be a Noetherian ring and I R an ideal. Then depth(i) codim(i). Proof. We have depth(i) = inf{depth(r p ) p V (I)} and codim(i) = inf{dim(r p ) p V (I)}, then the previous Proposition 1.58 implies the inequality. Finally, as one could expect there is a relation between the characterization of depth by means of the Koszul complex and by means of the Ext functor. Theorem 1.60. Let R be a ring, x = (x 1,..., x n ) a sequence in R, and M a finitely generated R-module. If I = (x) contains a weak M-sequence y = (y 1,..., y m ), then

1.5. Dimension and depth 30 H n+1 i K(x, M) = 0 for i = 1,..., m and H n m K(x, M) = Hom R (R/I, M/yM) = Ext m R (R/I, M). Proof. See Theorem 1.6.16, [6].

Chapter 2 Some fundamental theorems on syzygies In this chapter as the title says we will state and prove some interesting results in the theory of syzygies. To begin with, we will choose the Hilbert s syzygy theorem that was the starting point of this theory in his famous paper on Invariant Theory [7]. Secondly we shall prove a beautiful result conjectured by Serre: every finitely generated projective module over K[x 1,..., x n ] is free. Lastly we shall deal with the theory of our syzygies in the case of a local ring. 2.1 Towards Hilbert s syzygy theorem and more Lemma 2.1. If 0 A B C 0 is exact then pd(c) 1 + max(pd(a), pd(b)). Proof. If max(pd(a), pd(b)) is infinite we don t have anything to prove, therefore we assume it is finite. Let n = max(pd(a), pd(b)) then applying Theorem 1.19 for any module N we get the following piece of exact sequence... Ext n+1 (A, N) Ext n+2 (C, N) Ext n+2 (B, N)... 31

2.1. Towards Hilbert s syzygy theorem and more 32 Thus Theorem 1.43 and pd(a), pd(b) n implies... 0 Ext n+2 (C, N) 0... that is Ext n+2 (C, N) = 0. Finally by Theorem 1.43 we get pd(c) 1 + n. If M is an R-module then we denote M[x] = M R R[x] (2.1) as the R-module generated by the elements m R x k, where k N and m M, that is M[x] = M R x k k=0 which may be regarded as an R[x]-module taking into account the grading of the x k s. Also we have that if P is R-projective then P R R[x] is R[x]-projective, suppose we have the following diagram as in Definition 1.3, where B and C are R[x]-modules φ P R R[x] α B C 0. β Given the graded structure of P R R[x], it is enough to induce a function φ that accomplish α(p R x k ) = (β φ)(p R x k ) as an R-linear map and then we extend it to all of P R R[x] with the grading of the x k s. The isomorphism P R x k = P, the projectivity of P and looking at B and C as R-modules induces a map φ : P R R[x] B, where each restriction φ P R x k is an R-linear map that makes commute the previous diagram. Lemma 2.2. For every R-module M we have pd R[x] (M[x]) pd R (M). Proof. Suppose pd R (M) = n then there is an R-projective resolution d 0 P n d n... 2 d 1 d P1 0 P0 M 0. The ring of polynomials R[x] is a flat R-module (it is even free), then we can tensor

2.1. Towards Hilbert s syzygy theorem and more 33 R R[x] and obtain the exact sequence 0 P n R R[x] dn R1... d 2 R 1 P 1 R R[x] d 1 R 1 P 0 R R[x] d 0 R 1 M[x] 0. Where each P k R R[x] is an R[x]-projective module, then we get pd R[x] (M[x]) n. If M is an R[x]-module certainly we can see it as an R-module and we can also make the construction M[x] = M R R[x] that is an R[x]-module, where technically the action of x is given by x(m R x k ) = m R x k+1. Although xm is perfectly well defined, the operation x(m R x k ) = xm R x k is not correct for our construction 2.1. Lemma 2.3. If M is an R[x]-module then there is an R[x]-exact sequence 0 M[x] α M[x] β M 0. Proof. We define the map β by β(m R x k ) = x k m (M is already an R[x]-module) and the map α by α(m R x k ) = m R x k+1 xm R x k, for any k 0 and any m M. It is clear that β is surjective, for example restricting it we get β(m R 1) = M. If w = n k=0 m k R x k M[x] and α(w) = n k=0 m k R x k+1 n k=0 xm k R x k = xm 0 R 1 + n k=1 (m k 1 xm k ) R x k + m n R x n+1 = 0 then m 0 xm 1 = m 1 xm 2 =... = m n 1 xm n = m n = 0, so m 0 = m 1 =... = m n = 0 and α is injective. By the given definitions we have trivially β α = 0 then Im(α) Ker(β). But let w = n k=0 m k R x k Ker(β), hence β(w) = 0 and n k=0 x k m k = 0, which gives m 0 = n k=1 x k m k. The following elements u n 1 = m n u n 2 = m n 1 + xm n u n 3 = m n 2 + xm n 1 + x 2 m n. u 0 = m 1 + xm 2 +... + x n 1 m n

2.1. Towards Hilbert s syzygy theorem and more 34 have the property that n 1 n 1 α( u k R x k ) = xu 0 R 1 + (u k 1 xu k ) R x k + u n 1 R x n k=0 k=1 n = m k R x k. k=0 Therefore actually we have Im(α) = Ker(β). At this point we are finally ready to state and prove the main result of this section which relates the projective dimension over R[x] and over R. Theorem 2.4. Let M be an R[x]-module then pd R[x] (M) 1 + pd R (M). Proof. We take the previous R[x]-exact sequence 0 M[x] α M[x] β M 0, with Lemma 2.1 we find the inequality pd R[x] (M) 1 + pd R[x] (M[x]) and Lemma 2.2 gives pd R[x] (M[x]) pd R (M), which combined prove our assertion. Corollary 2.5. Let M be an R[x 1, x 2,..., x n ]-module then pd R[x1,x 2,...,x n](m) n + pd R (M). Proof. Make induction by successively taking R[x 1,..., x k 1, x k ] = R[x 1,..., x k 1 ][x k ]. Corollary 2.6. Let K be a field and M be a K[x 1, x 2,..., x n ]-module then pd K[x1,x 2,...,x n](m) n. Corollary 2.7. Let R be a ring then gdim(r[x 1, x 2,..., x n ]) n + gdim(r). Remark 2.8. We have achieved our goal of exposing a version of Hilbert s syzygy theorem in the most general settlement inside Commutative Algebra, i.e for commutative ring R; for a different approach see [3] in Chapter 19 and [5] in Chapter 3. Now unfortunately we will have to put additional properties to our ring R in order to get more interesting results. In the future we will stress which type of ring we are working and if it is not mentioned by